Leetcode 289. Game of Life

According to the Wikipedia’s article: “The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970.”

Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

1. Any live cell with fewer than two live neighbors dies, as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by over-population..
4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

Write a function to compute the next state (after one update) of the board given its current state. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input: 
[
  [0,1,0],
  [0,0,1],
  [1,1,1],
  [0,0,0]
]
Output: 
[
  [0,0,0],
  [1,0,1],
  [0,1,1],
  [0,1,0]
]

Follow up:

1. Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.
2. In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems?

解析：

一个m*n的单元格，每个单元格的初始状态为0或者1，分别表示live和dead，每个单元格的状态根据以下规则来确定：

1. 如果活细胞周围八个位置的活细胞数少于两个，则该位置活细胞死亡
2. 如果活细胞周围八个位置有两个或三个活细胞，则该位置活细胞仍然存活
3. 如果活细胞周围八个位置有超过三个活细胞，则该位置活细胞死亡
4. 如果死细胞周围正好有三个活细胞，则该位置死细胞复活

需要注意的是：所有单元格的状态需要同时更新，不能更新了部分单元格，这样会影响后面单元的状态的确定。并且要求in-place的方式，只能更新原有数组，并且需要知道未更新时的状态。这样考虑状态机来实现。

状态0： 死细胞转为死细胞

状态1： 活细胞转为活细胞

状态2： 活细胞转为死细胞

状态3： 死细胞转为活细胞

最后对所有状态对2取余，则状态0和2就变成死细胞，状态1和3就是活细胞，达成目的。先对原数组进行逐个扫描，对于每一个位置，扫描其周围八个位置，如果遇到状态1或2，就计数器累加1，扫完8个邻居，如果少于两个活细胞或者大于三个活细胞，而且当前位置是活细胞的话，标记状态2，如果正好有三个活细胞且当前是死细胞的话，标记状态3。完成一遍扫描后再对数据扫描一遍，对2取余变成我们想要的结果。

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int m = board.size();
        int n = m?board[0].size():0;
        vector<int> dx = {-1,-1,-1,0,1,1,1,0};
        vector<int> dy = {-1,0,1,1,1,0,-1,-1};
        for(int i=0; i < m; i++)
        {
            for(int j=0; j < n; j++)
            {
                int cnt = 0;
                for(int k=0; k <8; k++)
                {
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if(x >=0 && x < m && y>=0 && y <n && (board[x][y] == 1 || board[x][y] == 2) ){
                        cnt ++;
                    }
                        
                }
                if(board[i][j] && (cnt < 2 || cnt > 3))
                    board[i][j] = 2;
                else if(!board[i][j] && cnt == 3)
                    board[i][j] = 3;
            }
        }
        for(int i = 0; i < m; i++)
        {
            for(int j=0; j < n; j++)
            {
                board[i][j] %= 2;
            }
        }
        
    }
};

参考：https://www.cnblogs.com/grandyang/p/4854466.html